PSEUDODETERMINANTS AND PERFECT SQUARE SPANNING TREE COUNTS
JEREMY L. MARTIN, MOLLY MAXWELL, VICTOR REINER, AND SCOTT O. WILSON
Abstract. The pseudodeterminant pdet(M) of a square matrix is the last nonzero coefficient in its char- acteristic polynomial; for a nonsingular matrix, this is just the determinant. If ∂ is a symmetric or skew- symmetric matrix then pdet(∂∂t) = pdet(∂)2. Whenever ∂ is the kth boundary map of a self-dual CW- complex X, this linear-algebraic identity implies that the torsion-weighted generating function for cellular k-trees in X is a perfect square. In the case that X is an antipodally self-dual CW-sphere of odd dimension, the pseudodeterminant of its kth cellular boundary map can be interpreted directly as a torsion-weighted generating function both for k-trees and for (k − 1)-trees, complementing the analogous result for even- dimensional spheres given by the second author. The argument relies on the topological fact that any self-dual even-dimensional CW-ball can be oriented so that its middle boundary map is skew-symmetric.
1. Introduction This paper is about generating functions for higher-dimensional spanning tree via determinants of combi- natorial Laplacians, and when they are perfect squares. Combinatorial Laplacians, broadly interpreted, are matrices of the form L = ∂∂t where ∂ is a matrix in Zn×m, or perhaps even Rn×m where R is an integral domain containing Z along with indeterminates used as weights. Often L is singular, so that instead of considering the determinant det(L), one first creates an invertible reduced Laplacian by striking out some rows and columns. We will explore a useful alternative approach using the notion of pseudodeterminant [20, 26], appearing perhaps earliest in work of Adin [1, Theorem 3.4]. One first defines the rank r of L by extending scalars to the fraction field K of the domain R. Thus L will have r nonzero eigenvalues λ1, . . . , λr in the algebraic closure K of K, leading to two expansions for its unsigned characteristic polynomial:
X n−|I| n n−1 0 (1) det(t 1 +L) = t det(LI,I ) = t + trace(L)t + ··· + det(L)t I⊆{1,2,...,n} r Y = (t + λi)(2) i=1
where LI,I is the principal square submatrix of L indexed by row and column indices in the subset I. Definition 1.1. The pseudodeterminant pdet(L) is the last nonzero coefficient in the unsigned characteristic polynomial of L. That is, r X Y (3) pdet(L) := det(LI,I ) = λi. I⊆[n]: i=1 |I|=r Thus for nonsingular L one has pdet(L) = det(L), and when L is of rank one it has pdet(L) = trace(L). The pseudodeterminant is also the leading coefficient in the Fredholm determinant det(1 +tL) for the scaled Qn Qr r r−1 operator tL, that is, det(1 +tL) = i=0(1 + tλi) = i=0(1 + tλi) = pdet(L)t + O(t ); see, e.g., Simon [31, Chap. 3].
Date: January 2, 2015. 2010 Mathematics Subject Classification. 05C30, 05C05, 05E45, 15A15. Key words and phrases. pseudodeterminant, spanning tree, Laplacian, Dirac operator, perfect square, central reflex, self- dual. J.L. Martin was supported in part by a Simons Foundation Collaboration Grant and by National Security Agency grant no. H98230-12-1-0274. V. Reiner was supported by NSF grant DMS-1001933. 1 2 JEREMY L. MARTIN, MOLLY MAXWELL, VICTOR REINER, AND SCOTT O. WILSON
Section 2 quickly reviews and reformulates the well-known expansion of det(t 1 +L) via the Binet-Cauchy theorem. Section 3 digresses to explain a general perfect square phenomenon occurring if ∂2 = 0: (4) pdet(∂∂t) = pdet(∂ + ∂t)2. This happens because ∂ + ∂t plays the role of a combinatorial Dirac operator, whose square gives a sym- metrized Hodge-theoretic version of the combinatorial Laplacian: (5) (∂ + ∂t)2 = (∂2 + ∂∂t + ∂t∂ + (∂2)t) = ∂∂t + ∂t∂.
Section 4 proves a simple linear algebra result, Theorem 4.3, about the situation whenever ∂ in Zn×n is symmetric or skew-symmetric, that is, ∂t = ±∂: one then has (6) pdet(∂∂t) = (pdet ∂)2. When ∂ is the ith boundary map of a CW-complex X, the summands in (3) can be interpreted in terms of cellular spanning trees. This theory stems from the work of Kalai [19] and Bolker [7] and has been developed in recent work such as [1, 4, 10, 11, 12, 21, 24, 25]; we review it briefly in Section 5. In the special case that ∂ is symmetric or skew-symmetric, the linear-algebraic identity (6) says that the torsion-weighted number τi(X) of cellular spanning trees of X,
X 2 τi(X) := |H˜i−1(T )| , i-trees T ⊆X is a perfect square, as is a more general generating function for i- and (i − 1)-trees. The second author [25] explicated results of Tutte [32] and a question of Kalai [19, §7, Prob. 3], by showing that certain even-dimensional antipodally self-dual CW-spheres have spanning tree counts that are perfect squares, with a combinatorially significant square root. A goal of this paper is to prove similar results for odd-dimensional CW-spheres. In Section 6, we consider cellular d-balls S whose face posets are self-dual (a relatively weak self-duality condition), with no constraint on the parity of their dimension. Using Alexander duality, we show (Propo- sition 6.8) that τi(S) = τd−1−i(S), as well as an analogous formula for weighted tree counts. These results were observed by Kalai [19] for simplices, and our proof in the general case is based on Kalai’s ideas. A consequence of these results is that the pseudodeterminant of the (weighted or unweighted) middle Laplacian of a self-dual complex is always a perfect square (Corollary 6.10). In Section 7, we study the specific case of antipodally self-dual odd-dimensional spheres S =∼ S2k−1. Such t k a sphere can always be oriented so that the middle boundary map ∂ = ∂k satisfies ∂ = (−1) ∂. Together with Theorem 4.3, this implies that pdet(∂) has a direct combinatorial interpretation as τk(S) = τk−1(S); the weighted analogue of this statement is also valid (Theorem 7.4). The construction of the required orientation is technical and is deferred to an appendix (Section 8), although it can be made combinatorially explicit for certain spheres including polygons and boundaries of simplices.
2. Unsigned characteristic polynomials of Laplacians Let R be an integral domain, and let ∂ ∈ Rn×m (that is, ∂ is an n × m matrix over R). Let I ⊆ [n] := {1, 2, . . . , n} be a set of row indices and let J ⊆ [m] := {1, 2, . . . , m} be a set of column indices. The pair I,J |I|×|J| n×m determines an submatrix ∂I,J in R . Label row and column indices of ∂ in R with indeterminates x := (x1, . . . , xn) and y := (y1, . . . , ym). Let X = diag(x) be the square diagonal matrix having x as its diagonal entries, and likewise let Y = diag(y). For subsets I ⊆ [n] and J ⊆ [m], define monomials I Q J Q x := i∈I xi and y := j∈J yj. The following elementary proposition is well known. However, we were unable to find an explicit reference in the literature for its weighted version, so we include a proof for the sake of completeness.
Proposition 2.1. Every matrix ∂ ∈ Rn×m satisfies
t X n−|I| 2 (7) det(t 1 +∂∂ ) = t (det ∂I,J ) I⊆[n],J⊆[m]: |I|=|J| PSEUDODETERMINANTS AND PERFECT SQUARE SPANNING TREE COUNTS 3
1 t 1 n×n and more generally, the weighted operator L := X 2 ∂Y ∂ X 2 ∈ (R[x, y]) satisfies
X n−|I| I J 2 (8) det(t 1 +L) = t x y (det ∂I,J ) . I⊆[n],J⊆[m]: |I|=|J| In particular, if ∂ has rank r, then
X I J 2 (9) pdet(L) = x y (det ∂I,J ) . I⊆[n],J⊆[m]: |I|=|J|=r
1 1 1 t 1 t Proof. Let Z := X 2 ∂Y 2 , so that L := X 2 ∂Y ∂ X 2 = ZZ . The Binet-Cauchy identity gives
X t det LI,I = det ZI,J det ZJ,I J⊆[m]: |J|=|I| and the principal minor expansion (1) gives
X n−|I| X n−|I| X t det(t 1 +L) = t det LI,I = t det ZI,J det ZJ,I I⊆[n] I⊆[n] J⊆[m]: |J|=|I| X 1 1 1 1 n−|I| I 2 J 2 I 2 t J 2 = t (x ) · det ∂I,J · (y ) (y ) · det ∂I,J · (x ) I⊆[n],J⊆[m]: |J|=|I| X n−|I| I J 2 = t x y (det ∂I,J ) , I⊆[n],J⊆[m]: |J|=|I|
proving (8). Setting xi = yj = 1 for all i, j recovers (7). To obtain (9), note that R[x, y] is an integral t domain, and ∂, ∂∂ ,L all have rank r.
The nonzero summands in (9) are those for which ∂I,J is nonsingular. Accordingly, we can reformulate the summation indices in (9). Definition 2.2. For a domain R and ∂ in Rn×m, say that a subset of row indices I ⊆ [n] forms a row basis for ∂ if, after extending scalars to the fraction field K of R, the rows of ∂ indexed by I give a K-vector space basis for the row space of ∂. Similarly define for J ⊆ [m] what it means to be a column basis for ∂. We will write RowB(∂) and ColB(∂) for the set of row and column bases, respectively. Proposition 2.3. (cf. [2, Chap. 4, Exer. 2.5]) Let R be a domain and ∂ in Rn×m of rank r, and let I ⊆ [n], J ⊆ [m] have |I| = |J| = r. Then the submatrix ∂I,J is nonsingular if and only if both I is a row basis and J is a column basis for ∂.
J I Proof. Extending scalars from R to K, factor the K-linear map ∂I,J : K → K into ∂I,J = β ◦ α as follows:
∂I,J
∂ π ' KJ / Km / Kn / / KI < i
" " - im ∂
α β Kr / Kr / Kr In the top row, the first horizontal inclusion KJ ,→ Km pads a vector in KJ with extra zero coordinates m n I outside of J to create a vector in K , while the last horizontal surjection K K forgets the coordinates outside of I. The factorization ∂I,J = β ◦ α shows that ∂I,J is nonsingular if and only if both α and β are nonsingular, that is, if and only if both J is a column basis and I is a row basis for ∂. 4 JEREMY L. MARTIN, MOLLY MAXWELL, VICTOR REINER, AND SCOTT O. WILSON
Proposition (2.3) allows us to rewrite equation (9) as follows:
X I J 2 (10) pdet(L) = x y (det ∂I,J ) . I∈RowB(∂) J∈ColB(∂)
3. Digression: Pseudodeterminants and Laplacians as squares of Dirac operators This section will not be used in the sequel. We first collect a few easy properties of pseudodeterminants analogous to properties of determinants, then apply them to show why the pseudodeterminant of the Dirac operator ∂ + ∂t, defined for ∂ in Zn×n satisfying ∂2 = 0, agrees up to ± sign with the pseudodeterminant for either of the Laplace operators ∂∂t or ∂t∂. As usual, R will be a domain, with fraction field K having algebraic closure K. Proposition 3.1. (cf. Knill [20, Prop. 2]) For L in Rn×n, one has the following. (a) pdet(Lt) = pdet(L). (b) pdet(Lk) = pdet(L)k for k = 1, 2,.... (c) If A, B lie in Rn×m,Rm×n, respectively, then pdet(AB) = pdet(BA). (d) If L, M in Rn×n are mutually annihilating (i.e., LM = 0 = ML), then pdet(L + M) = pdet(L) pdet(M). Proof. Assertion (a) follows from det(t 1 +Lt) = det ((t 1 +L)t) = det(t 1 +L). k Assertion (b) follows since if L has nonzero eigenvalues λ1, . . . , λr, then L has nonzero eigenvalues k k λ1 , . . . , λr . Assertion (c) comes from a well-known determinant fact (see, e.g., [30]) asserting that, if n ≥ m, then det(t 1 +AB) = tn−m det(t 1 +BA). n Assertion (d) will follow by making a change of coordinates in K that simultaneously triangularizes the mutually annihilating (and hence mutually commuting) matrices L, M. Thus without loss of generality, L and M are triangular and have as their ordered lists of diagonal entries their eigenvalues (λ1, . . . , λn), (µ1, . . . , µn). Their sum L + M is then triangular, with eigenvalues (λ1 + µ1, . . . , λn + µn). However, since either product ML or LM is also triangular, with eigenvalues (λ1µ1, . . . , λnµn), the mutual annihilation 0 = LM = ML implies that at most one of each pair {λi, µi} can be nonzero. Hence if L, M have ranks r, s, respectively, then the eigenvalues for L + M can be reindexed as (λ1, . . . , λr, µr+1, . . . , µr+s, 0, 0,..., 0) and hence
pdet(L + M) = λ1 ··· λr · µr+1 ··· µr+s = pdet(L) pdet(M).
Definition 3.2. For ∂ in Zn×n with ∂2 = 0, its Dirac operator is the symmetric matrix ∂ + ∂t. The reader is referred to Friedrich [15] for background on Dirac operators in Riemannian geometry. As noted in equation (5) in the Introduction, this operator ∂ + ∂t has square given by ∆ := (∂ + ∂t)2 = (∂2 + ∂∂t + ∂t∂ + (∂2)t) = ∂∂t + ∂t∂ which is another form of combinatorial Laplacian, arising in discrete Hodge theory over R; see, e.g., Friedman [14]. The subspace of harmonics H := ker ∆ ⊆ Rn gives a canonical choice of representatives for the homology ker ∂/ im ∂, due to the orthogonal Hodge decomposition picture:
Rn = im ∂t ⊕ H ⊕ im ∂ ker ∂ = ker ∂t∂ = H ⊕ im ∂ ker ∂t = ker ∂∂t = im ∂t ⊕ H ker ∂ ∩ ker ∂t = H.
Corollary 3.3. Let ∂ in Zn×n be such that ∂2 = 0. Then its Dirac operator ∂ + ∂t satisfies
t t X 2 pdet(∂ + ∂ ) = ± pdet(∂∂ ) = ± (det ∂I,J ) . I∈RowB(∂) J∈ColB(∂) PSEUDODETERMINANTS AND PERFECT SQUARE SPANNING TREE COUNTS 5
Proof. The second equality comes from specializing all variables to 1 in the right-hand side of (10). For the first equality, we check that its two sides have the same square: 2 pdet(∂ + ∂t) = pdet (∂ + ∂t)2 (Prop. 3.1(b)) = pdet ∂∂t + ∂t∂ (Eqn. (5)) = (pdet ∂∂t)(pdet ∂t∂) (Prop. 3.1(d)) = pdet(∂∂t)2. (Prop. 3.1(c)) For the third equality, note that ∂∂t and ∂t∂ are mutually annihilating: (∂t∂)(∂∂t) = ∂t(∂2)∂t = 0 and t t 2 t (∂∂ )(∂ ∂) = ∂(∂ ) ∂ = 0. Remark 3.4. All four operators ∂∂t, ∂t∂, ∂∂t + ∂t∂, and ∂ + ∂t are self-adjoint. The first three are positive semidefinite, and hence have nonnegative pseudodeterminant by Proposition 3.1(c). However, the Dirac 0 1 operator ∂ + ∂t can be indefinite and have negative pseudodeterminant. For example, ∂ = has 0 0 0 1 ∂2 = 0, and its Dirac operator ∂ + ∂t = has eigenvalues (+1, −1), with pdet(∂ + ∂t) = −1. 1 0
4. Symmetry or skew-symmetry Something interesting happens to the pseudodeterminant in our previous results when ∂ happens to be square and either symmetric or skew-symmetric, due to the following fact. Lemma 4.1. For R a domain, with ∂ in Rn×m of rank r, and any r-subsets A, A0 ⊆ [n] and B,B0 ⊆ [m],
det ∂A,B det ∂A0,B0 = det ∂A0,B det ∂A,B0 . In particular, when n = m, for any r-subsets I,J of [n] one has
det ∂I,I det ∂J,J = det ∂I,J det ∂J,I . Proof. Extend scalars to the fraction field K of R. Considering ∂ as a K-linear map Km −→ Kn, its rth exterior power is a K-linear map r ∧rKm −−→∧ ∂ ∧rKn. m n r m r n If K ,K have standard bases (v1, . . . , vm) and (w1, . . . , wn), then ∧ K , ∧ K have K-bases of wedges
vA := va1 ∧ · · · ∧ var and wB := wb1 ∧ · · · ∧ wbr indexed by r-subsets A ⊆ [m] and B ⊆ [n]. The matrix for r r ∧ ∂ in these bases has (A, B)-entry det(∂A,B). To prove the lemma, it suffices to show that ∧ ∂ has rank 1, so its 2 × 2 minors vanish. k r m n Let us show more generally that ∧ ∂ has rank k . Make changes of bases in K ,K via invertible matrices P,Q in GLm(K), GLn(K) , so that ∂ = PDQ, where I 0 D = r . 0 0 Then ∧k∂ = ∧kPDQ = ∧kP · ∧kD · ∧kQ, where ∧kP, ∧kQ have inverses ∧k(P −1), ∧k(Q−1), and " # k I r 0 ∧ D = (k) 0 0